CBSE Class 11 Mathematics: Relations and Functions NCERT Solutions

We are given two ordered pairs that are equal: $\backslash left(\backslash frac\{x\}\{3\}+1,\; y-\backslash frac\{2\}\{3\}\backslash right)=\backslash left(\backslash frac\{5\}\{3\},\; \backslash frac\{1\}\{3\}\backslash right)$. For two ordered pairs to be equal, their corresponding elements must be equal. Therefore, we can set up two separate equations:

1. Equating the first elements: $\backslash frac\{x\}\{3\}\; +\; 1\; =\; \backslash frac\{5\}\{3\}$
2. Equating the second elements: $y\; -\; \backslash frac\{2\}\{3\}\; =\; \backslash frac\{1\}\{3\}$

Now, we solve each equation for its respective variable:

For the first equation (x):

\frac{x}{3} = \frac{5}{3} - 1

\frac{x}{3} = \frac{5}{3} - \frac{3}{3}

\frac{x}{3} = \frac{2}{3}

Multiplying both sides by 3, we get:

x = 2

For the second equation (y):

y = \frac{1}{3} + \frac{2}{3}

y = \frac{3}{3}

y = 1

Thus, the values are x=2 and y=1.

We are given that set A has 3 elements. Let n(A) denote the number of elements in set A, so n(A) = 3.

We are also given set B as B = \{3, 4, 5\}. The number of elements in set B, denoted by n(B), is 3.

The Cartesian product of two sets A and B, denoted by A \times B, is the set of all possible ordered pairs (a, b) such that a \in A and b \in B. The number of elements in the Cartesian product A \times B is given by the product of the number of elements in each set:

n(A \times B) = n(A) \times n(B)

Substituting the given values:

n(A \times B) = 3 \times 3

n(A \times B) = 9

Therefore, the number of elements in (A \times B) is 9.

We are given the sets G = \{7, 8\} and H = \{5, 4, 2\}.

The Cartesian product G \times H is the set of all ordered pairs (g, h) where g \in G and h \in H.

To find G \times H, we pair each element of G with each element of H:

G \times H = \{(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)\}

Similarly, the Cartesian product H \times G is the set of all ordered pairs (h, g) where h \in H and g \in G.

To find H \times G, we pair each element of H with each element of G:

H \times G = \{(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)\}

1. False. The correct statement is: If $P\; =\; \backslash \{m,\; n\backslash \}$ and $Q\; =\; \backslash \{n,\; m\backslash \}$, then $P\; \backslash times\; Q\; =\; \backslash \{(m,\; n),\; (m,\; m),\; (n,\; n),\; (n,\; m)\backslash \}$. Explanation: The Cartesian product $P\; \backslash times\; Q$ includes all possible ordered pairs. We pair each element of P with each element of Q. So, $m$ from P pairs with $n$ and $m$ from Q, giving $(m,\; n)$ and $(m,\; m)$. Similarly, $n$ from P pairs with $n$ and $m$ from Q, giving $(n,\; n)$ and $(n,\; m)$.
2. True. Explanation: By definition, the Cartesian product $A\; \backslash times\; B$ of two non-empty sets A and B is the set of all ordered pairs $(x,\; y)$ where $x$ is an element of A and $y$ is an element of B. Since A and B are non-empty, there will be at least one such ordered pair, making $A\; \backslash times\; B$ non-empty.
3. True. Explanation: The intersection of any set with the empty set ($\backslash Phi$) is always the empty set. So, $B\; \backslash cap\; \backslash Phi\; =\; \backslash Phi$. The Cartesian product of any set with the empty set is also the empty set. Therefore, $A\; \backslash times\; \backslash Phi\; =\; \backslash Phi$.